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đặt \(\left\{{}\begin{matrix}S=X+Y\\P=X.Y\end{matrix}\right.\)
a)\(\left\{{}\begin{matrix}S+P=5\\S^2-P=7\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}P=5-S\\S^2+S-12=0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}P=5-S\\\left[{}\begin{matrix}S=-4\\S=3\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}S=-4\\P=9\end{matrix}\right.\\\left\{{}\begin{matrix}S=3\\P=2\end{matrix}\right.\end{matrix}\right.\)
suy ra tìm đc x và y
b,c tương tự
do \(x=0\) và \(y=0\) không phải là một nghiệm của hệ nên
HPT\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1+x^3y^3}{x^3}=\dfrac{19x^3}{x^3}\\\dfrac{y+xy^2}{x^2}=-\dfrac{6x^2}{x^2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x^3}+y^3=19\\\dfrac{y}{x^2}+\dfrac{y^2}{x}=-6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(\dfrac{1}{x}+y\right)^3-\dfrac{3y}{x}\left(\dfrac{1}{x}+y\right)=19\\\dfrac{y}{x}\left(\dfrac{1}{x}+y\right)=-6\end{matrix}\right.\)
Đặt \(\dfrac{1}{x}+y=u\) ; \(\dfrac{y}{x}=v\)
HPT\(\Leftrightarrow\left\{{}\begin{matrix}u^3-3uv=19\\uv=-6\end{matrix}\right.\)
\(\Rightarrow u^3+18=19\Rightarrow u^3=1\)\(\Rightarrow u=1\)
\(\Rightarrow v=-6\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+y=1\\\dfrac{y}{x}=-6\end{matrix}\right.\)
\(\Rightarrow6x^2+x-1=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
đến đây thì ez rồi
a: \(\Leftrightarrow\left\{{}\begin{matrix}35x-28y=21\\35x-45y=40\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}17y=-19\\5x-4y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{19}{17}\\x=-\dfrac{5}{17}\end{matrix}\right.\)
b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{8}{y}=18\\\dfrac{10}{x}+\dfrac{8}{y}=102\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{11}{x}=120\\\dfrac{1}{x}-\dfrac{8}{y}=18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{11}{120}\\y=-\dfrac{44}{39}\end{matrix}\right.\)
c: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{30}{x-1}+\dfrac{3}{y+2}=3\\\dfrac{25}{x-1}+\dfrac{3}{y+2}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{x-1}=1\\\dfrac{10}{y-1}+\dfrac{1}{y+2}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=5\\\dfrac{1}{y+2}+2=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=6\\y=-3\end{matrix}\right.\)
d: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{135}{2x-y}+\dfrac{160}{x+3y}=35\\\dfrac{135}{2x-y}-\dfrac{144}{x+3y}=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+3y=8\\2x-y=9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+6y=16\\2x-y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=5\end{matrix}\right.\)
a, \(x\in\left[-3;2\right]\)
b, \(x\in[0;5)\)
c, \(x\in\varnothing\)